| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2020 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Tank/reservoir mixing problems |
| Difficulty | Standard +0.3 This is a straightforward integrating factor question with standard steps: rearrange to standard form, find integrating factor 1/(1+t), integrate to find general solution, apply initial condition, then evaluate at specific times. The algebra is routine and the context is typical. Slightly easier than average (0.0) as it follows a mechanical procedure with no conceptual surprises, though the square root term requires careful integration. |
| Spec | 4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((1+t)\frac{dP}{dt}+P = t^{\frac{1}{2}}(1+t) \Rightarrow \frac{dP}{dt}+\frac{P}{1+t}=t^{\frac{1}{2}}\) | B1 | Correct rearrangement; may be implied by subsequent work |
| \(I = e^{\int\frac{1}{1+t}dt} = 1+t \Rightarrow P(1+t) = \int t^{\frac{1}{2}}(1+t)\,dt = \ldots\) | M1 | Uses model to find integrating factor and attempts solution |
| \(P(1+t) = \frac{2}{3}t^{\frac{3}{2}}+\frac{2}{5}t^{\frac{5}{2}}+c\) | A1 | Correct solution |
| \(t=0,\ P=5 \Rightarrow c=5\) | M1 | Interprets initial conditions to find constant; must use \(t=0\) and \(P=5\) |
| \(P = \frac{\frac{2}{3}t^{\frac{3}{2}}+\frac{2}{5}t^{\frac{5}{2}}+5}{(1+t)} = \frac{\frac{2}{3}(8^{\frac{3}{2}})+\frac{2}{5}(8^{\frac{5}{2}})+5}{9} = \ldots\) | M1 | Uses solution to find population after 8 hours |
| \(= 10\,277\) bacteria (allow awrt 10 300) | A1 | cso; correct number from correct equation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P = \frac{\frac{2}{3}t^{\frac{3}{2}}+\frac{2}{5}t^{\frac{5}{2}}+5}{(1+t)} \Rightarrow \frac{dP}{dt} = \frac{(1+t)(t^{\frac{1}{2}}+t^{\frac{3}{2}})-(\frac{2}{3}t^{\frac{3}{2}}+\frac{2}{5}t^{\frac{5}{2}}+5)}{(1+t)^2}\) | M1 | Differentiates model using appropriate method; trivialised differentiation from incorrect work is M0 |
| Correct differentiation of correct answer to (a) up to constant of integration | A1ft | Follow through on their \(c\); if implicit differentiation used must get to function in \(t\) only |
| \(\left(\frac{dP}{dt}\right)_{t=1} = \frac{5\times10-\left(\frac{16}{3}+\frac{64}{5}+5\right)}{(5)^2} = \frac{403}{375}\) | dM1 | Uses \(t=4\) in their \(\frac{dP}{dt}\) to obtain value |
| \(\frac{403}{375}\times 1000 = \frac{3224}{3}\) (= awrt 1070) bacteria per hour | A1 | Correct answer; allow 1075 or answers rounding down to 1070 with correct units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P = \frac{\frac{2}{3}t^{\frac{3}{2}}+\frac{2}{5}t^{\frac{5}{2}}+5}{(1+t)} = \frac{\frac{16}{3}+\frac{64}{5}+5}{(1+4)}\) | M1 | Substitutes \(t=4\) into their \(P\) |
| \(= \frac{347}{75}\) | A1ft | Correct value for \(P\); follow through on their constant from (a) |
| \((1+t)\frac{dP}{dt}+P = t^{\frac{1}{2}}(1+t) \Rightarrow 5\frac{dP}{dt}+\frac{347}{75} = 2\times5 \Rightarrow \frac{dP}{dt}=\frac{403}{375}\) | dM1 | Uses \(t=4\) and their \(P\) to find \(\frac{dP}{dt}\) |
| \(\frac{403}{375}\times1000 = \frac{3224}{3}\) (= 1075) bacteria per hour | A1 | Correct answer with correct units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| E.g. The number of bacteria increases indefinitely which is not realistic | B1 | Suggests suitable limitation referring to the model; allow sensible comment even without equation. Do not allow answers referring to external factors such as temperature |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(1+t)\frac{dP}{dt}+P = t^{\frac{1}{2}}(1+t) \Rightarrow \frac{dP}{dt}+\frac{P}{1+t}=t^{\frac{1}{2}}$ | B1 | Correct rearrangement; may be implied by subsequent work |
| $I = e^{\int\frac{1}{1+t}dt} = 1+t \Rightarrow P(1+t) = \int t^{\frac{1}{2}}(1+t)\,dt = \ldots$ | M1 | Uses model to find integrating factor and attempts solution |
| $P(1+t) = \frac{2}{3}t^{\frac{3}{2}}+\frac{2}{5}t^{\frac{5}{2}}+c$ | A1 | Correct solution |
| $t=0,\ P=5 \Rightarrow c=5$ | M1 | Interprets initial conditions to find constant; must use $t=0$ and $P=5$ |
| $P = \frac{\frac{2}{3}t^{\frac{3}{2}}+\frac{2}{5}t^{\frac{5}{2}}+5}{(1+t)} = \frac{\frac{2}{3}(8^{\frac{3}{2}})+\frac{2}{5}(8^{\frac{5}{2}})+5}{9} = \ldots$ | M1 | Uses solution to find population after 8 hours |
| $= 10\,277$ bacteria (allow awrt 10 300) | A1 | cso; correct number from correct equation |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P = \frac{\frac{2}{3}t^{\frac{3}{2}}+\frac{2}{5}t^{\frac{5}{2}}+5}{(1+t)} \Rightarrow \frac{dP}{dt} = \frac{(1+t)(t^{\frac{1}{2}}+t^{\frac{3}{2}})-(\frac{2}{3}t^{\frac{3}{2}}+\frac{2}{5}t^{\frac{5}{2}}+5)}{(1+t)^2}$ | M1 | Differentiates model using appropriate method; trivialised differentiation from incorrect work is M0 |
| Correct differentiation of correct answer to (a) up to constant of integration | A1ft | Follow through on their $c$; if implicit differentiation used must get to function in $t$ only |
| $\left(\frac{dP}{dt}\right)_{t=1} = \frac{5\times10-\left(\frac{16}{3}+\frac{64}{5}+5\right)}{(5)^2} = \frac{403}{375}$ | dM1 | Uses $t=4$ in their $\frac{dP}{dt}$ to obtain value |
| $\frac{403}{375}\times 1000 = \frac{3224}{3}$ (= awrt 1070) bacteria per hour | A1 | Correct answer; allow 1075 or answers rounding down to 1070 with correct units |
**Alternative for (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P = \frac{\frac{2}{3}t^{\frac{3}{2}}+\frac{2}{5}t^{\frac{5}{2}}+5}{(1+t)} = \frac{\frac{16}{3}+\frac{64}{5}+5}{(1+4)}$ | M1 | Substitutes $t=4$ into their $P$ |
| $= \frac{347}{75}$ | A1ft | Correct value for $P$; follow through on their constant from (a) |
| $(1+t)\frac{dP}{dt}+P = t^{\frac{1}{2}}(1+t) \Rightarrow 5\frac{dP}{dt}+\frac{347}{75} = 2\times5 \Rightarrow \frac{dP}{dt}=\frac{403}{375}$ | dM1 | Uses $t=4$ and their $P$ to find $\frac{dP}{dt}$ |
| $\frac{403}{375}\times1000 = \frac{3224}{3}$ (= 1075) bacteria per hour | A1 | Correct answer with correct units |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| E.g. The number of bacteria increases indefinitely which is not realistic | B1 | Suggests suitable limitation referring to the model; allow sensible comment even without equation. Do not allow answers referring to external factors such as temperature |\begin{enumerate}
\item A sample of bacteria in a sealed container is being studied.
\end{enumerate}
The number of bacteria, $P$, in thousands, is modelled by the differential equation
$$( 1 + t ) \frac { \mathrm { d } P } { \mathrm {~d} t } + P = t ^ { \frac { 1 } { 2 } } ( 1 + t )$$
where $t$ is the time in hours after the start of the study.\\
Initially, there are exactly 5000 bacteria in the container.\\
\begin{enumerate}[label=(\alph*)]
\item Determine, according to the model, the number of bacteria in the container 8 hours after the start of the study.
\item Find, according to the model, the rate of change of the number of bacteria in the container 4 hours after the start of the study.
\item State a limitation of the model.
\end{enumerate}
\hfill \mbox{\textit{Edexcel CP1 2020 Q7 [11]}}